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5 GmatPrep questions

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Manager
Joined: 13 Jan 2009
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5 GmatPrep questions [#permalink]  13 Aug 2009, 04:45
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Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
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GmatPrep Q's.doc [242.5 KiB]

Manager
Joined: 25 Jul 2009
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Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 05:10
Try uploading them as jpegs instead of docs.....that way members will be able to view the question in their explorer windows itself....
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Senior Manager
Joined: 25 Jun 2009
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Kudos [?]: 87 [0], given: 6

Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 05:41
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!

Q 1 .See the attached figure.

Let the sides of the triangle be a, b and c
Clearly using simple trigo calculations we can see that a=b= 5
and c= 5\sqrt{2}
Now lets draw perpendicular on C with the height h.
This perpendicular will bisect the side C,
So let X= 1/2 x C = 1/2 x 5\sqrt{2}

h^2 + x^2 = a ^2
H^2 = 25- 25/2 = 25/2 = 5/ \sqrt{2}

Area = base x height = 1/2 C x H = 1/2 x 5\sqrt{2} x 5/ \sqrt{2}
Area = 25/2 = 12.5
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untitled.JPG [ 30.92 KiB | Viewed 1414 times ]

Senior Manager
Joined: 25 Jun 2009
Posts: 309
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Kudos [?]: 87 [0], given: 6

Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 05:47
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!

Q 2 See the attached figure.

Distance traveled = D = 2 pi R = 2pi = 2 x 3.14

where r = radius of the lake = 1 mile

Distance = speed x time
Speed = 3 miles per hour
time = Distance / speed =(2 x 3.14) / 3

Clearly t >2 and less than 2.5

Hence Option C is the answer.
Attachments

untitled.JPG [ 28.33 KiB | Viewed 1406 times ]

Senior Manager
Joined: 25 Jun 2009
Posts: 309
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Kudos [?]: 87 [0], given: 6

Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 05:49
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!

2^x - 2^(x-2) = 3 ( 2^13)
2^(x-2) [ 4-1]= 3 ( 2^13)
Hence x-2= 13
OR X= 15
Attachments

untitled.JPG [ 28.95 KiB | Viewed 1409 times ]

Manager
Joined: 25 Jul 2009
Posts: 117
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
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Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 05:54
SOL for Q10)

Method 1: There is a coordinate geometry formula that uses matrices to give you the area of a triangle whose coordinates are known.
The final formula is: If the coordinates of a triangle are A(x1,y1), B(x2,y2) & C(x3,y3), then the area of the triangle is given by,
1/2* [(x1-x2)(y2-y3) - (y1-y2)(x2-x3)]

Applying we get,
A(PQR) = 1/2* [(4-0)(3-4) - (0-3)(0-7)]
= 1/2 [(-4) - (21)]
= 1/2 [-25] .......... ignore the '-' sign
= 12.5
ANS: A

Note: This method is also useful to decide whether the given 3 points form a triangle or they are collinear. If the area turns out to be zero, then the points are collinear.

Method 2: Drop perpendiculars from point R to the X and Y axes at points S(7,0) and T(0,4) respectively. Thus OSRT is a rectangle.

Now A(PQR) = A(OSRT) - A(OQP) - A(QRT) - A(PRS) where all the three triangles are right triangles.
A(PQR) = 7*4 - (3*4/2) - (7*1/2) - (3*4/2)
= 12.5
ANS: A

Method 3: Find out the lengths of the three sides using distance formula and then use the formula that calculates the area of a triangle from its sides.
_________________

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My GMAT Debrief - 740 (Q50, V39) | My Test-Taking Strategies for GMAT | Sameer's SC Notes

Senior Manager
Joined: 25 Jun 2009
Posts: 309
Followers: 2

Kudos [?]: 87 [0], given: 6

Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 06:02
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!

The way u can represent the scores = 48, 48 , 70, 70, 70 , 80 , 80, 80, 80, 84 , 84 , 84, 84, 84, 84 , 84, 96, 96, 96,96

So clearly the Median = 84

See the figure for this question .

I don't know how to work on the 5th question may be some body can shed some light on this one.
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untitled.JPG [ 30.62 KiB | Viewed 1396 times ]

Manager
Joined: 25 Jul 2009
Posts: 117
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
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Kudos [?]: 172 [0], given: 17

Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 06:04
SOL for Q24

Its given that the student worked for 20 days and earned:
48$- on 2 days 70$ - on 3 days
80$- on 4 days 84$ - on 7 days
96\$ - on 4 days

For Median of even no of amounts, we need to arrange the amounts in ascending order and take the avg of the amounts placed in 10th and 11th positions => (84 + 84)/2 = 84

ANS: B
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My GMAT Debrief - 740 (Q50, V39) | My Test-Taking Strategies for GMAT | Sameer's SC Notes

Senior Manager
Joined: 25 Jun 2009
Posts: 309
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Kudos [?]: 87 [0], given: 6

Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 06:05
samrus98 wrote:
SOL for Q10)

Method 1: There is a coordinate geometry formula that uses matrices to give you the area of a triangle whose coordinates are known.
The final formula is: If the coordinates of a triangle are A(x1,y1), B(x2,y2) & C(x3,y3), then the area of the triangle is given by,
1/2* [(x1-x2)(y2-y3) - (y1-y2)(x2-x3)]

Applying we get,
A(PQR) = 1/2* [(4-0)(3-4) - (0-3)(0-7)]
= 1/2 [(-4) - (21)]
= 1/2 [-25] .......... ignore the '-' sign
= 12.5
ANS: A

Note: This method is also useful to decide whether the given 3 points form a triangle or they are collinear. If the area turns out to be zero, then the points are collinear.

Method 2: Drop perpendiculars from point R to the X and Y axes at points S(7,0) and T(0,4) respectively. Thus OSRT is a rectangle.

Now A(PQR) = A(OSRT) - A(OQP) - A(QRT) - A(PRS) where all the three triangles are right triangles.
A(PQR) = 7*4 - (3*4/2) - (7*1/2) - (3*4/2)
= 12.5
ANS: A

Method 3: Find out the lengths of the three sides using distance formula and then use the formula that calculates the area of a triangle from its sides.

Hey Sameer,

I was actually wondering if there is a faster way of solving this one and u just came up with that ..

Cheers
Manager
Joined: 25 Jul 2009
Posts: 117
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
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Kudos [?]: 172 [0], given: 17

Re: 5 GmatPrep questions [#permalink]  13 Aug 2009, 06:13
SOL for Q32)

Mean = 8.1
SD = 0.3

Now we need to calculate the acceptable range for the observations measuring 1.5 SDs on either side of the mean => 8.1 + (0.3)*1.5 => 8.1 + 0.45
=> 7.65 to 8.55

Except for 7.51 all the entries are within 1.5 SDs of the mean.

ANS: E
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Manager
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Re: 5 GmatPrep questions [#permalink]  27 Aug 2009, 15:27
nitishmahajan wrote:
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!

2^x - 2^(x-2) = 3 ( 2^13)
2^(x-2) [ 4-1]= 3 ( 2^13)
Hence x-2= 13
OR X= 15

how did you get what's in red?

thanks
Re: 5 GmatPrep questions   [#permalink] 27 Aug 2009, 15:27
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